Two Dimensional Vectors, level 12.

In this video we will go over how to solve static equilibrium problems. In general when

an object is stationary or at rest we say it is in a state of static equilibrium. When

an object is in this state, the sum of the forces acting on the object must be equal

to 0. For example say we had an object sitting on

a table. In its current position the trophy is not moving but, if the table is suddenly

and rapidly removed, the trophy will fall to the ground and break. What force caused

the trophy to fall down immediately after removing the table? The force responsible

is the force of gravity. This force is perhaps the most intuitive force in our lives since

it pulls us towards the center of the earth on a daily basis. The magnitude of this force

is our weight which is the product of our mass and the gravitational acceleration; the

direction of this force is always towards the center of the earth. Let’s go ahead

and place the trophy back on the table. Since the trophy has a mass it will experience a

gravitational force equal to its mass (m) times the gravitational acceleration (g) pushing

it down towards the center of the earth. Once again this is nothing more than the objects

weight. Now the reason why the trophy does not break the table and move towards the center

of the earth is because the table is pushing the trophy in the opposite direction, and

the magnitude of this force just happens to be the same as the force of gravity exerted

on the trophy, hence the resultant force is equal to zero and the trophy remains motionless. When an object is in a state of static equilibrium

the sum of the forces acting on the object must be equal to 0. In other words the resultant

force equals 0. These types of problem are solved more extensively and in greater detail

in a typical physics course. For this multivariable calculus course we will go over 3 basic examples.

Naveen weighs 42.0 pounds. He is sitting on a swing when his sister Vianey pulls him and

the swing back horizontally through an angle of 30 degrees and then stops. Find the tension

in the ropes of the swing and the magnitude of the force exerted by Vianey.

Alright the figure of this situation is shown below. Notice that Naveen is motionless at

the moment that Vianey stops swinging back. At this point, Vianey is pulling with enough

force to keep Naveen in a state of static equilibrium. In addition to Vianey’s force,

Naveen is under the influence of 2 other forces the gravitational force and the tension force

on the ropes. We are asked to find the magnitude of this tension force and the magnitude of

the force exerted by Vianey. Similar to the way we solved resultant force problems we

need to find the component form of all the forces exerted on the object then we go ahead

and add them component wise and finally we take this sum and set it equal to the zero

vector. Alright let’s go head and first find the component form of all the forces

acting on Naveen. Using the geometry of the diagram we can draw

a right triangle with the ropes representing the hypotenuse and the x and y components

of the tension force representing the legs. For this problem you are free to use the 30

degree angle or the 60 degree angle that is formed by the sides of the right triangle

as long as you apply the trigonometric ratios correctly and assign the correct direction

to the x and y component, everything will work out. I am going to use the 60 degree

angle and use it to find the components of the tension force. In this case the x-component

of the tension force will be equal to the magnitude of the tension force which is currently

not known, times cosine of 60 degrees and the y-component will be equal to the magnitude

of the tension force times sine of 60 degrees. Alright now let’s find the components of

the gravitational force acting on Naveen. Since the gravitational force is always pointing

towards the center of the earth it will only have one component in this case the vertical

or y-component and its magnitude will be equal to Naveen’s weight. Also notice that it

will be negative since it is pointing in the direction of the negative y-axis. So the

components of the gravitational force are going to be equal to 0 for the x-component

and negative 42 pounds for the y-component. Now let’s find an expression for the components

of Vianey’s force, similar to the gravitational force this force will only have one component

which is facing towards the positive x-axis, so the component form of vianey’s force

will be equal to the magnitude of vianey’s pulling force (which is currently not known),

for the x-component and 0 for the y-component. Now that we have all 3 forces we go ahead

and add the forces component wise. Also since this is a static equilibrium problem we also

need to set it equal to the 0 vector as follows. This expression allows us to generate 2 separate

equations one for the x-component and one for the y-component. In essence we are dealing

with a system of equations in this case we have 2 unknowns which require 2 equations

to solve and we just happen to have those equations. So now let’s go ahead and solve

for the tension force with the second equation, approximating the expression we obtain about

48.5 pounds for the tension force. Lastly we use this value and the first equation to

solve and approximate Vianey’s force which is approximately equal to 24.3 pounds. Alright,

let’s try the next example. The tension at each end of the chain has a

magnitude of 25 newtons. What is the weight of the chain? Assume that the weight of the

chain is concentrated at the midpoint of the chain.

Alright here we have a hanging chain under tension forming an angle of 37 degrees with

the horizontal portion of the edges. We are also advised to assume that the weight of

the chain is located at the midpoint of the chain. This means that we are going to have

3 distinct forces acting on the chain focused at the midpoint. Similar to the previous example let’s first

find an expression for the components of all three forces. For the first tension force

the x-component will be equal to the magnitude of the tension force in this case 25 newtons

times cosine of 37 degrees and this component will be negative since it is pointing in the

direction of the negative x-axis. The y-component on the other hand will be equal to 25 newtons

times sine of 37 degrees and it will be pointing towards the positive y-axis. For the second

tension force the magnitudes of both components will be the same, but the x-component will

be pointing towards the positive x-axis. Lastly the gravitational force will be pointing

directly down towards the center of the earth. So its components will be equal to 0 for the

x-component and negative times the magnitude of the weight which is currently unknown.

Next we go ahead and add these 3 vectors and set them equal to the zero vector. Now we

are able to generate our 2 equations by adding the x-components together and the y-components

together, doing that we obtain the following equations. Notice that the first equation

represents the sum of all the x-components acting on the chain and it is composed of

only two forces in this case the x-component of each of the tension force, since this problem

is highly symmetrical we expect these two forces to cancel out and this equation confirms

that observation. The second equation tells us that the entire weight of the chain is

being held by the vertical components of the tension forces, so we go ahead and solve for

the weight, doing that and approximating the expression we obtain 30.1 newtons for the

weight of the chain. Alright let’s go over the final example.

Use the figure below to determine the tension in each cable supporting the given load

Similar to the previous examples let’s start by finding the components of each force, the

first tension force will have the following components, and the second tension force components

will be equal to the following. Finally the gravitational force will have the following

values for its components. Notice that both tension forces are unknowns,

since this problem lacks symmetry we need to get a little bit more involved in order

to find the values of these forces. Let’s go ahead and add these forces and set the

expressions equal to the zero vector. From here let’s extract our two equations

one from each component as follows. Now we are face to face with a system of equations

with 2 unknowns. At this point it is just a matter of solving for one of the unknown

tensions from one equation and substituting it back into the second equation and solving

for the unknown variables. So solving for the first tension force we obtain the following.

Then we go ahead and substitute this expression into the first equation as follows, we proceed

by solving for the magnitude of the second force which is approximately equal to 1958.1

pounds, from here we use this value and solve for the second tension force which is approximately

equal to 2638.2 pounds. Alright in our next video we will go over distance and bearing

problems.

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I do no 12 shit your not my dad you little stupid turtle fu…;-)

lolðŸ‘»ðŸ˜‚ðŸ˜• what

fuck the math coming straight from the underground bitch